Math - More about Equations (Logarithm)?
回答 (1)
2020-09-18 1:44 pm
✔ 最佳答案
log_a(ab) + log_b(ab) + log_a(a-b) + log_b(a-b) = 0= log_a(ab(a-b))+log_b(ab(a-b))
= (log_a(b)+1)log_b(ab(a-b))
∴ log_a(b) = -1, or ab(a-b) = 1
log_a(b) = -1 iff. (if and only if) ab = 1.
ab(a-b) = 1
iff. a^2 b - ab^2 - 1 = 0
iff. a = b + √(b^4+4b)/(2b) (∵ a > b > 0)
收錄日期: 2021-04-24 08:00:31
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20200917230139AA9QK4R